The light of two different frequencies whose photons have energies...

Expression for maximum kinetic energy can be written as, {\left(KE\right)}_\max = E - ϕ

${\left(KE\right)}_{max, 1} = {E}_{1} - ϕ = 3 . 8 - 0 . 6 = 3 . 2eV$, ${\left(KE\right)}_{max, 2} = {E}_{2} - ϕ = 1 . 4 - 0 . 6 = 0 . 8eV$

$\dfrac{{\left(KE\right)}_{max, 1}}{{\left(KE\right)}_{max, 2}} = \dfrac{3 . 2}{0 . 8} = 4 ⇒ 4 = \dfrac{\dfrac{1}{2}{mv}_{1}^{2}}{\dfrac{1}{2}m{v}_{2}^{2}}$

$⇒ {v}_{1}^{2}{v}_{2}^{2} = 4$ $⇒ \dfrac{{v}_{1}}{{v}_{2}} = 2: 1$